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[DP] Add a solution to Wildcard Matching
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/** | ||
* Question Link: https://leetcode.com/problems/wildcard-matching/ | ||
* Primary idea: Classic Two Dimensionel Dynamic Programming | ||
* Time Complexity: O(mn), Space Complexity: O(mn) | ||
*/ | ||
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class WildcardMatching { | ||
func isMatch(_ s: String, _ p: String) -> Bool { | ||
let sChars = Array(s), pChars = Array(p) | ||
var dp = Array(repeating: Array(repeating: false, count: p.count + 1), count: s.count + 1) | ||
dp[0][0] = true | ||
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// must start from 0, to make range feasible and handle empty vs. * case | ||
for i in 0...s.count { | ||
for j in 0...p.count { | ||
guard j > 0 else { | ||
continue | ||
} | ||
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let pCurrent = pChars[j - 1] | ||
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if pCurrent != "*" { | ||
dp[i][j] = i > 0 && dp[i - 1][j - 1] && (pCurrent == sChars[i - 1] || pCurrent == "?") | ||
} else { | ||
// Two situations: | ||
// (1) '*' is the first character in p; | ||
// (2) For k>=0 and k<=i, there is some dp[k][j-1] being true; | ||
// and '*' will match the rest sequence in s after index k; | ||
var flag = false | ||
for k in 0...i { | ||
if dp[k][j - 1] { | ||
flag = true | ||
break | ||
} | ||
} | ||
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dp[i][j] = flag || j == 1 | ||
} | ||
} | ||
} | ||
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return dp[s.count][p.count] | ||
} | ||
} |
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